This fails already for $d=3$.

Consider a tetrahedron, e.g. the convex hull of the points $v_1,v_2,v_3,v_4$. Let $K$ be the closed subset consisting of $\sum_{i=1}^4 a_i v_i$ with $\sum_{i=1}^4 a_i=1$ and $a_3 +a_4 \leq 1/2-\epsilon$. Let $L$ be defined similarly, but $a_1+a_2 \leq 1/2-\epsilon$. Clearly the convex hull of $K$ and $L$ is this tetrahedron.

Consider a shift vector $x$ which lowers the volume of the convex hull. By the convexity described in Alexandre's answer, we can assume $x$ is invariant under all symmetries of the situation. By the symmetry switching $v_1$ and $v_2$ and the one switching $v_3$ and $v_4$, $x$ is a multiple of $v_1+v_2-v_3-v_4$. It suffices to handle the case where it is a small positive multiple. (For a negative multiple, there would just be a local minimum further out.)

The chance in volume for small $x$ is the sum of, for each triangular face of the tetrahedron, the integral of the dot product of the surface normal of the face with any vector describing the change in the boundary of the polyhedron. The dot product of the surface normal of each face with $x$ is the same, up to sign. If we normalize so this is $1$, then this dot product will lie between $0$ and $1$ on the faces spanned by $v_1,v_2,v_3$ and $v_1,v_2,v_4$, and between $0$ and $-1$ on the spaces spanned by $v_1,v_3,v_4$ and $v_2,v_3,v_4$.

On the face $v_1,v_2,v_3$, this is the maximal convex function that is $0$ at $v_1$ and $v_2$, $1$ on $L$ (which are the points $av_1+bv_2+cv_3$ with $c \geq 1/2 +\epsilon$). In particular, with $c < 1/2+\epsilon$, its value is $c/(1/2+\epsilon)$. Clearly the integral depends continuously on $\epsilon$, so we will evaluate in the case $\epsilon=0$. So the integral of this function divided by the area of the face is $$\frac{ \int_0^{1/2} (1-x) (2x) dx + \int_{1/2}^1 (1-x)dx }{ \int_0^1 (1-x) dx} = \frac{ 1/4 - 1/12 +1/2 - 3/8 }{1/2} = \frac{7}{12}$$

The same is true for the face $v_1,v_2,v_4$.

On the face $v_2,v_3,v_4$, this is the maximal concave function that is $-1$ on $v_3$ and $v_4$ and $0$ whenever $a \geq 1/2+\epsilon$, in other words when $a < 1/2+\epsilon$ it is $ -( (1/2+\epsilon)-a )/(1/2+\epsilon)$. Again there is a continuity and we can take $\epsilon=0$. Then the integral is

$$ - \frac{ \int_{0}^{1/2} (1-x) (1-2x) dx } { \int_0^{1} (1-x) dx} =+ \frac{1/2 -3/8 + 1/12}{1/2}= - \frac{5}{12}$$

Because $\frac{7}{12}-\frac{5}{12}>0$, the change in the $x$ direction is positive, and it remains so for $\epsilon$ sufficiently small.